Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.

Consider the function f defined by f(x)=x. (Let's keep it as simple as possible). If you pick a small number epsilon, I can always (no matter what number you picked) find a positive number delta such that |f(x)-0| is less than epsilon for all x such that |x-0| is less than delta. That's what we mean when we say that the function goes to 0 as x goes to 0.

The point is that by choosing x "close enough" to 0 (the value that x goes to), we can make f(x) be "close enough" to 0 (the limit of f(x)).

f(x) approaches a limit L (f(x) -> L) as x approaches a (x -> a) if we can make f(x) as close to L as we want provided that x is sufficiently close, but unequal to, a. This is probably the intuitive notion you are familiar with. Note we are not concerned with whether f is even defined at a since we only care about behavior of f as x gets arbitrarily close to a.

Let's be very specific to begin with. It is not hard to convince yourself that the function f(x) = [tex]\sqrt{|x|}cos(x)[/tex] approaches 0 as x approaches 0. We want to make f(x) close to 0, so why not make f(x) within [tex]\frac{1}{10}[/tex] of 0?

Now we need to make x sufficiently close to a.
By inspection, if we make x within [tex]\frac{1}{100}[/tex] of 0, i.e.,
[tex]|x| < \frac{1}{100}[/tex],
then [tex]|\sqrt{|x|}cos(x)| \leq \sqrt{|x|} < \sqrt{\frac{1}{100}} = \frac{1}{10}[/tex].

What epsilon was used in the above example? Could we have chosen any number to be epsilon? Which delta was chosen and how does it relate to the given epsilon? If you can relate each component of the specific example shown with those of the precise definition, then you prove the more general statement (bolded above).

That's definitely not an intuitive or simple definition. In one sentence, you have some of the hardest parts of logic put together: mixed quantifiers and implication (and the implication is often written "backwards" using the phrasing "whenever"). But nevermind all the details. Just get a feel for it. You never need the definition ever again once you've proven the continuity of the important classes functions. But it's a good workout for the brain!

In real life, you can never have perfect precision. In any measurement you make, there will always be some amount of error. If you are careful, you can make the bounds of that error very small.

In the definition of a limit, epsilon and delta are error bounds. That is why the absolute values are there. When you say [tex] |x - a| < \delta[/tex], what you mean is that x and a are equal within an error tolerance of delta. Similarly with epsilon, f(x), and L.

A continuous function has to do with accuracy of input and output of the function. There needs to be a relationship between accurate output and accurate input. More specifically, a continuous function must be one that allows you get an output as accurate as you desire.... simply by inputting accurate enough data.

When coming up with proofs of continuity, you are usually looking for a value for delta in terms of epsilon that satisfy the logical implication. These proofs also make heavy use of the triangle rule for absolute value.

I still don't get it very nicely. I would appreciate if you could put it in simpler words.

The [tex]\epsilon - \delta[/tex] definition of limit is given as follows:

Let [tex]f(x)[/tex] be defined in the neighbourhood of [tex]a[/tex]. Then [tex]f(x)[/tex] is said to tend to limit [tex]L[/tex] as [tex]x[/tex] approaches to [tex]a[/tex], or symbolically,
[tex]lim_{x \rightarrow a} f(x) =L[/tex]
If for every positive number [tex]\epsilon[/tex] - however small it be - there exists a corresponding positive number [tex]\delta[/tex] such that
[tex]0 < |x - a| < \delta \mbox{then} |f(x) - L| < \epsilon[/tex]

I have some questions:
1. Can [tex]x[/tex] be equal to [tex]a[/tex]?
2. Is is that we can select [tex]\epsilon[/tex] and [tex]\delta[/tex] ourselves? How? Can any small number serve as [tex]\epsilon[/tex]?

I am also confused about the "Existence of a funstion at a point" part:

Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.

Since 1/0.0009=1111.111..., the smallest integer we can use for this purpose is 1112. So I pick N=1112, and you say "ha-ha, that N is useless when I choose epsilon to be 0.000007 instead". But then I calculate 1/0.000007=142857.142... and choose N=142858.

The meaning of the statement "the limit of this sequence is 0" is that I will always win this game if I get to choose my number (N) after you choose yours (epsilon).

I have some questions:
1. Can [tex]x[/tex] be equal to [tex]a[/tex]?
2. Is is that we can select [tex]\epsilon[/tex] and [tex]\delta[/tex] ourselves? How? Can any small number serve as [tex]\epsilon[/tex]?

1. The function f doesn't have to be defined at a, so if we allow x=a, the inequality [itex]|f(x)-L|<\epsilon[/itex] doesn't always make sense. So [itex]0<|x-a|<\delta[/itex] is right.

([itex]\forall[/itex] means "for all". [itex]\exists[/itex] means "there exists"). The statement "there exists a number [itex]\delta>0[/itex] such that [itex]0<|x-a|<\delta\implies |f(x)-y|<\epsilon[/itex]" is going to be true when [tex]\epsilon=6\cdot 10^{11}[/itex] for most functions f that you will encounter. But this tells you nothing, since we only say that f(x)→y when x→a if that statement holds for all [itex]\epsilon>0[/itex], including every member of (for example) the sequence 1/2n.

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?

You probably just heard him wrong or misunderstood him. Consider the example [itex]f(x)=\sqrt x[/itex] and choose [itex]\epsilon=\frac 1 2[/itex]. We obviously want to define the limit in a way that means that this f(x)→0 as x→0, but do you think it's possible to chose a [itex]\delta\geq\epsilon=\frac 1 2[/itex] such that [itex]0<|x-0|<\delta \implies |f(x)-0|<\epsilon[/itex]? When x is slightly less than 1/2, f(x) is approximately 1.4 which is [itex]>\epsilon=0.5[/itex], so [itex]|f(x)-0|<\epsilon[/itex] doesn't hold.

So if I pick epsilon=0.00001, what would delta be? Say [tex]lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0 [/tex]. What would be delta? How do we find out? How are epsilon and delta related to each other? So delta is not smaller than or equal to epsilon?

This is how my book solves this question:
Find the limit [tex]lim_{x \rightarrow 3} (3x - 4)[/tex] and verify the result.
Solution:
[tex]lim_{x \rightarrow 3} (3x - 4)[/tex] = 3*3-4 = 5. To verify that result, we have to show that the corresponding to [tex]\epsilon > 0[/tex], there exists [tex]\delta[/tex] such that [tex]|(3x - 4) - 5| < \epsilon[/tex] whenever [tex]|x-3| < \delta[/tex].

Here we have to find [tex]\delta[/tex] in terms of [tex]\epsilon[/tex], considering [tex]\epsilon[/tex] is given. Now

Fred and George are going to play a game. The game goes as follows:
(1) Fred writes a positive number. We will call that number [itex]\epsilon[/itex].
(2) George writes a positive number. We will call that number [itex]\delta[/itex].
(3) Fred writes down another number. We will call that number [itex]x[/itex].

The definition of limit says that if [itex]\lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0[/itex], then there is a strategy George can use that will let him win every time. If that limit doesn't exist or is not zero, then there is a strategy Fred can use that will let him win every time.

How is delta found out in terms of epsilon in this solution? Is |x - 3| equal to delta? How? I didn't get this solution. Please help me out.

What part of it are you having problems with? The solution you posted clearly shows that if we choose δ=ε/3 (or smaller), we have |(3x-4)-5|<ε for all x that satisfies 0<|x-3|<δ, and by definition of the limit, that means that 3x-4→5 as x→3.

I don't see why you're asking if |x-3| is equal to delta. It seems that you have ignored the words "for all" in all of the replies you got. You should think about what they mean.

Obviously the limit is 5, i.e. [tex]\lim_{x \rightarrow 5} x = 5[/tex] but how do you prove
that? (It is obvious if you draw it. Do you know the geometric meaning of the limit?)

To show that the limit is 5, you have to show the following:
[tex]|x-a|< \delta \Rightarrow |f(x)-L| < \epsilon[/tex]
(this is just taken from the definition. L is the limit)

In other words:Proof:
Step 0. Choose an appropriate value for delta
Step 1. You start with [tex]|x-a|< \delta[/tex],
Step 2. then make some manipulations and
Step 3. arrive at [tex]|f(x)-L| < \epsilon[/tex]End of proof
--

In our example [tex]\lim_{x \rightarrow 5} x[/tex],

we have a=5 and f(x)=x and we want
to show that L=5. For this, we have to show that:

Step 1. You start with [tex]|x-5|< \delta[/tex],
Step 2. then make some manipulations
Step 3. and arrive at [tex]|x-5|< \epsilon[/tex].

How do you get from Step 1 to Step 3, i.e. from [tex]|x-5|< \delta[/tex] to [tex]|x-5|< \epsilon[/tex]?
See Step 0: Choose an appropriate value for delta.
An obvious choice is delta = epsilon.
Let's check that:

Anyone notice that when functions are not simple linear expressions, that the epsilon-delta limit proofs are usually not easy?

That's because linear functions are among the simplest of functions! That's why things like derivatives and the mean value theorem are so important -- they allow you to approximate complicated functions with linear functions. Another particular example is the limit of a quadratic expression; one controls a quadratic term like

[tex]|x^2 - a|[/tex]​

by factoring it into a product of linear terms

[tex]|(x - \sqrt{a})| |(x + \sqrt{a}) |[/tex]​

and then finding a way to simultaneously control both (linear) factors.